1 mole of NH3 gas at 27oC is expanded adiabatic condition to make volume 8 times - -1-33- Find temperature and work done respectively-

The correct option is A 150 K, 900 cal TV^γ 1= Constant #160; #160; #160; #160; #160;[Adiabatic condition] (T_2)(V_2)^0.33=( ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Relation between P, V, T, Gamma in Adiabatic Proceses

1 mole of N...

Question

$1$ mole of $N H_{3}$ gas at $27^{o}$ C is expanded adiabatic condition to make volume $8$ times $(γ = 1.33)$ . Find temperature and work done respectively.

150

900

cal

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

150

400

cal

No worries! We‘ve got your back. Try BYJU‘S free classes today!

250

1000

cal

No worries! We‘ve got your back. Try BYJU‘S free classes today!

200

800

cal

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $150$ K, $900$ cal
$T V^{γ - 1} =$ Constant [Adiabatic condition]
$(T_{2}) (V_{2})^{0.33} = (T_{1}) (V_{1})^{0.33}$
$V_{2} = 8 V_{1}$
$∴ T_{2} = \frac{T_{1}}{(8)^{0.33}} = \frac{300}{(8)^{0.33}} \approx 150 K$
$W = \frac{n R (T_{2} - T_{1})}{γ - 1}$
$= - \frac{1 \times 2 \times 150}{0.33} C a l \approx 900 C a l$
Work done $= 900 C a l$

Suggest Corrections

Similar questions

Q. 1 mole of

N H_{3}

gas at

27^{o} C

is expanded in reversible adiabatic condition to make volume 8 times

(γ = 1.33) .

Final temperature and magnitude of work done respectively are

Q. 1 mole of

N H_{3}

gas at

27^{o} C

is expanded in reversible adiabatic condition to make volume 64 times

(γ = 1.33) .

Final temperature and magnitude of work done respectively are

1

mole of

N H_{3}

gas at

27^{o}

C is expanded adiabatic condition to make volume

8

times

(γ = 1.33)

. Final temperature and work done respectively are:

1

mole of

N H_{3}

gas at

27^{\circ} C

is expanded in reversible adiabatic condition to make the volume

8

times

(γ = 1.33)

. Final temperature and work done respectively are:

1

mole of

N H_{3}

gas at

27^{\circ} C

is expanded in reversible adiabatic condition to make volume

8

times

(γ = 1.33)

. Calculate the final temperature and work done respectively.

Join BYJU'S Learning Program

1 mole of NH3 gas at 27oC is expanded adiabatic condition to make volume 8 times (γ=1.33). Find temperature and work done respectively.

The correct option is A 150 K, 900 calTVγ−1= Constant [Adiabatic condition](T2)(V2)0.33=(T1)(V1)0.33V2=8V1∴T2=T1(8)0.33=300(8)0.33≈150KW=nR(T2−T1)γ−1=−1×2×1500.33Cal≈900CalWork done =900Cal

$1$ mole of $N H_{3}$ gas at $27^{o}$ C is expanded adiabatic condition to make volume $8$ times $(γ = 1.33)$ . Find temperature and work done respectively.